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            <div class="post-toc animated"><ol class="nav"><li class="nav-item nav-level-1"><a class="nav-link" href="#%E4%B8%89%E5%88%86"><span class="nav-number">1.</span> <span class="nav-text">三分</span></a><ol class="nav-child"><li class="nav-item nav-level-2"><a class="nav-link" href="#%E7%A4%BA%E4%BE%8B"><span class="nav-number">1.1.</span> <span class="nav-text">示例</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#%E4%BE%8B%E4%B8%89%E5%88%86%E5%87%BD%E6%95%B0"><span class="nav-number">1.1.1.</span> <span class="nav-text">例：三分|函数</span></a></li></ol></li></ol></li></ol></div>
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    <div class="post-body" itemprop="articleBody"><h1 id="三分">三分</h1>
<p>三分是一种用于求解<strong>单峰/单谷函数</strong>极值的搜索算法，通过每次将区间分成三份并比较两个分点的函数值来缩小搜索范围。</p>
<span id="more"></span>
<p>对于一个单峰函数，在区间 <span class="math inline">\([l,r]\)</span>
上有唯一的极大值点。可以取两个三分点：</p>
<ul>
<li><span class="math inline">\(mid1 = l + \frac{r-l}{3}\)</span></li>
<li><span class="math inline">\(mid2 = r - \frac{r-l}{3}\)</span></li>
</ul>
<img src="/2025-04-02-19-%E4%B8%89%E5%88%86/trisection.svg" class="">
<p>比较 <span class="math inline">\(f(mid1)\)</span> 和 <span
class="math inline">\(f(mid2)\)</span> 的大小:</p>
<ul>
<li>如果 <span class="math inline">\(f(mid1) &lt;
f(mid2)\)</span>，说明极大值点在 <span
class="math inline">\([mid1,r]\)</span> 区间内，令 <span
class="math inline">\(l = mid1\)</span></li>
<li>如果 <span class="math inline">\(f(mid1) &gt;
f(mid2)\)</span>，说明极大值点在 <span
class="math inline">\([l,mid2]\)</span> 区间内，令 <span
class="math inline">\(r = mid2\)</span></li>
<li>如果 <span class="math inline">\(f(mid1) =
f(mid2)\)</span>，说明极大值点在 <span
class="math inline">\([mid1,mid2]\)</span> 区间内，令 <span
class="math inline">\(l = mid1, r = mid2\)</span></li>
</ul>
<p>每次迭代可以将区间长度缩小为原来的 <span
class="math inline">\(\frac{2}{3}\)</span>。</p>
<p>对于单谷函数，只需要将比较符号反向即可求出极小值点。</p>
<p>三分的基本框架如下:</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 三分求单峰函数极大值点</span></span><br><span class="line"><span class="function"><span class="type">double</span> <span class="title">TrisectionMax</span><span class="params">(<span class="type">double</span> l, <span class="type">double</span> r)</span> </span>&#123;</span><br><span class="line">    <span class="type">double</span> mid1, mid2;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">100</span>; i ++) &#123;</span><br><span class="line">        mid1 = l + (r - l) / <span class="number">3</span>;</span><br><span class="line">        mid2 = r - (r - l) / <span class="number">3</span>;</span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">f</span>(mid1) &lt; <span class="built_in">f</span>(mid2)) l = mid1;</span><br><span class="line">        <span class="keyword">else</span> r = mid2;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> l;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 三分求单谷函数极小值点</span></span><br><span class="line"><span class="function"><span class="type">double</span> <span class="title">TrisectionMin</span><span class="params">(<span class="type">double</span> l, <span class="type">double</span> r)</span> </span>&#123;</span><br><span class="line">    <span class="type">double</span> mid1, mid2;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">100</span>; i ++) &#123;</span><br><span class="line">        mid1 = l + (r - l) / <span class="number">3</span>;</span><br><span class="line">        mid2 = r - (r - l) / <span class="number">3</span>;</span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">f</span>(mid1) &gt; <span class="built_in">f</span>(mid2)) l = mid1;</span><br><span class="line">        <span class="keyword">else</span> r = mid2;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> l;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="示例">示例</h2>
<h3 id="例三分函数">例：三分|函数</h3>
<p>给定 <span class="math inline">\(n\)</span> 个二次函数 <span
class="math inline">\(f_1(x), f_2(x), \ldots, f_n(x)\)</span>（均形如
<span class="math inline">\(ax^2 + bx + c\)</span>），设 <span
class="math inline">\(F(x) = \max\{f_1(x), f_2(x), \ldots,
f_n(x)\}\)</span>，求 <span class="math inline">\(F(x)\)</span> 在区间
<span class="math inline">\([0, 1000]\)</span> 上的最小值。</p>
<p>输入</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">2</span><br><span class="line">1</span><br><span class="line">2 0 0</span><br><span class="line">2</span><br><span class="line">2 0 0</span><br><span class="line">2 -4 2</span><br></pre></td></tr></table></figure>
<p>输出</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">0.0000</span><br><span class="line">0.5000</span><br></pre></td></tr></table></figure>
<p>分析：都是朝上的二次函数，画画图可以发现，两个口朝上的单峰函数取较大函数值叠加，仍然是口朝上的单峰函数，那么无数个叠加也仍然是，可以用三分求极值。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;algorithm&gt;</span></span></span><br><span class="line"></span><br><span class="line"><span class="type">const</span> <span class="type">int</span> maxn = <span class="number">1e4</span> + <span class="number">10</span>;</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> n;</span><br><span class="line"></span><br><span class="line"><span class="keyword">struct</span> <span class="title class_">Func</span> &#123;</span><br><span class="line">    <span class="type">double</span> a, b, c;</span><br><span class="line">    <span class="function"><span class="type">double</span> <span class="title">calc</span><span class="params">(<span class="type">double</span> x)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> a * x * x + b * x + c;</span><br><span class="line">    &#125;</span><br><span class="line">&#125; f[maxn];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">double</span> <span class="title">F</span><span class="params">(<span class="type">double</span> x)</span> </span>&#123;</span><br><span class="line">    <span class="type">double</span> ret = f[<span class="number">0</span>].<span class="built_in">calc</span>(x);</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">1</span>; i &lt; n; i++)</span><br><span class="line">        ret = std::<span class="built_in">max</span>(ret, f[i].<span class="built_in">calc</span>(x));</span><br><span class="line">    <span class="keyword">return</span> ret;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">double</span> <span class="title">TrisectionMin</span><span class="params">(<span class="type">double</span> l, <span class="type">double</span> r)</span> </span>&#123;</span><br><span class="line">    <span class="type">double</span> mid1, mid2;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">100</span>; i ++) &#123;</span><br><span class="line">        mid1 = l + (r - l) / <span class="number">3</span>;</span><br><span class="line">        mid2 = r - (r - l) / <span class="number">3</span>;</span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">F</span>(mid1) &gt; <span class="built_in">F</span>(mid2)) l = mid1;</span><br><span class="line">        <span class="keyword">else</span> r = mid2;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> l;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="type">int</span> T;</span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;T);</span><br><span class="line">    <span class="keyword">while</span>(T--) &#123;</span><br><span class="line">        <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;n);</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++)</span><br><span class="line">            <span class="built_in">scanf</span>(<span class="string">&quot;%lf%lf%lf&quot;</span>, &amp;f[i].a, &amp;f[i].b, &amp;f[i].c);</span><br><span class="line">            </span><br><span class="line">        <span class="built_in">printf</span>(<span class="string">&quot;%.4f\n&quot;</span>, <span class="built_in">F</span>(<span class="built_in">TrisectionMin</span>(<span class="number">0</span>, <span class="number">1000</span>)));</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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